## Chapter 3

## Q. 3.22

The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure:

N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)

If 18.20 kg of NH_{3} is produced by a reaction mixture that initially contains 6.00 kg of H_{2} and an excess of N_{2}, what is the percent yield of the reaction?

## Step-by-Step

## Verified Solution

**Collect and Organize** We know that the actual yield of NH_{3} is 18.20 kg. We also know that the reaction mixture initially contained 6.00 kg of H_{2} and that H_{2} must be the limiting reactant because the problem specifies the presence of excess N_{2}. We need to find the percent yield.

**Analyze** We need to use the mass of H_{2} to calculate how much NH_{3} could have been produced— the theoretical yield. We can then use the theoretical yield and the actual yield to calculate percent yield. We need the molar masses of H_{2} and NH_{3} and the stoichiometry of the reaction, which tells us that 2 moles of NH_{3} are produced for every 3 moles of H_{2} consumed.

**Solve** The molar masses we need are:

\mathscr{M}_{H_{2}} = 2(1.008 g/mol) =2.016 g/mol

\mathscr{M}_{NH_{3}} =14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol

We calculate the theoretical yield of NH_{3}:

6.00 \sout{kg H_{2}}\times \frac{10^{3} \sout{g H_{2}}}{1 \sout{kg H_{2}}} \times \frac{1 \sout{mol H_{2}}}{2.016 \sout{g H_{2}}} \times \frac{2 \sout{mol NH_{3}}}{3 \sout{mol H_{2}}} \times \frac{17.03 \sout{g NH_{3}}}{1 \sout{mol NH_{3}}} \times \frac{1 kg NH_{3}}{10^{3} \sout{g NH_{3}}} =33.8 kg NH_{3}

Then we divide the actual yield by this theoretical yield to determine the percent yield:

\frac{18.20 \sout{kg NH_{3}}}{33.8 \sout{kg NH_{3}}} \times 100 \%=53.8\%

**Think About It** A yield of about 54% may seem low, but it may be the best that can be achieved for a particular process. A great deal of chemical research goes into trying to improve the percent yield of industrial chemical reactions.